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Formula is C10H14

  1. Aug 31, 2004 #1
    This should be easy, but I am missing something.

    5 hydrogens at 7 ppm seeing what appears to be an overlap of several
    1 hydrogen at 2.5 ppm seeiing 5
    2 hydrogen at 1.5 ppm seeing 4
    3 hydrogen at 1.23 ppm seeing 1
    3 hydrogen at .75ppm seeing 2

    The formula is C10H14

    It appears I have a benzene ring but I am unable to get the side chain, which I am asumming there is only 1 and with the 3 seeing 2 and 2 seeing 3, there should be a CH2CH3.

  2. jcsd
  3. Aug 31, 2004 #2


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    Your notation is not familiar with me, but I will try to predict the formula from your data.

    1) The 7-ppm-centered peak system is due to a benzene ring, a plain phenyl substituent causes a characteristic multiplet. So we have a C6H5 for now.

    2) A pentet at 2,5 ppm (I think you want to mean this by saying "seeing 5") having one proton means there is a branching CH group, one methyl and one ethyl; a total of five protons seen with only one proton on hand is possible this way.

    3) A quartet at 1,5 ppm having two protons indicate a CH2-CH3 group (on the left).

    4) A singlet peak at 1,23 ppm with three protons clearly shows that this group is a CH3.

    5) A doublet at 0,75 ppm with three protons is indicative of CH2-CH3 group (on the right).

    So the molecule you are seeking is methylcumene, C6H5-CH(CH3)-CH2-CH3, which can far more easily seen from the attachment .zip file containing a tif image file.

  4. Aug 31, 2004 #3
    Great, thanks. That is what I had drawn, but for some reason it seemed like something wasn't right.

    thanks again
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